Metrically bounded set

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"Bounded set" is misused, there are other notions of bounded perhaps of sets. As such metrically bounded (set) ought to be used. This is the start of that process Alec (talk) 23:10, 18 March 2017 (UTC)


Let [ilmath](X,d)[/ilmath] be a metric space and let [ilmath]A\in\mathcal{P}(X)[/ilmath] be an arbitrary subset of [ilmath]X[/ilmath]. We say [ilmath]A[/ilmath] is (metrically)[Note 1] bounded or is a (metrically) bounded set if[1]:

  • [ilmath]\exists C\in[/ilmath][ilmath]\mathbb{R}_{>0} [/ilmath][ilmath]\forall a,b\in A[d(a,b)<C][/ilmath]
    • In words: there exists a real [ilmath]C>0[/ilmath] such that for any two points in [ilmath]A[/ilmath] the distance between them (as measured by the metric, [ilmath]d[/ilmath]) is strictly less than [ilmath]C[/ilmath]

Note that if [ilmath]A\eq\emptyset[/ilmath] then it is vacuously seen to be bounded by any [ilmath]C>0[/ilmath][Note 2]

Equivalent statements

Implying statements

See also


  1. Alec's terminology to create a "global name" for use with pages
  2. Proof:
    • Define (choose) [ilmath]C:\eq 1\in\mathbb{R} [/ilmath]
      • By rewriting for-all and exists within set theory we see:
        • [ilmath]\forall a,b\in A[d(a,b)<C][/ilmath] is short for [ilmath]\forall a\in A\forall b\in A[d(a,b)<C][/ilmath] which is short for [ilmath]\forall a[a\in A\implies \forall b[b\in A\implies d(a,b)<C]][/ilmath]
        • It is easy to see this is equivalent to [ilmath]\forall a\forall b[(a\in A\wedge b\in A)\implies d(a,b)<C][/ilmath]
        • Let [ilmath]a,b[/ilmath] be given
          • As [ilmath]A\eq\emptyset[/ilmath] we cannot have [ilmath]a\in A[/ilmath] or [ilmath]b\in A[/ilmath] so the LHS of the implication is false
            • By the nature of logical implication we consider it true regardless of the RHS when the LHS is false, so we're done.


  1. Functional Analysis - Volume 1: A gentle introduction - Dzung Minh Ha